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Tangents and Normals

CSEC Additional Mathematics Essential Knowledge: Tangents and normals are fundamental concepts in calculus and coordinate geometry. A tangent is a straight line that touches a curve at a single point without crossing it, while a normal is perpendicular to the tangent at that point. These concepts have applications in physics, engineering, and optimization problems.

Key Concept: For a curve \(y = f(x)\), at any point \(P(x_1, y_1)\) on the curve:

Tangent: A line that touches the curve at point P with gradient equal to \(f'(x_1)\) (the derivative at that point)
Normal: A line perpendicular to the tangent at point P, with gradient \(-\frac{1}{f'(x_1)}\) (negative reciprocal of tangent’s gradient)

Part 1: Fundamental Concepts and Relationships

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What Are Tangents and Normals?

Tangent Line

• Touches curve at one point
• Gradient = \(f'(x_1)\)
• “Just touches” the curve

Normal Line

• Perpendicular to tangent
• Gradient = \(-\frac{1}{f'(x_1)}\)
• Passes through same point

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The Perpendicular Relationship

The most important relationship between tangents and normals:

\[m_{\text{tangent}} \times m_{\text{normal}} = -1\]

Or equivalently:

\[m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\]

Where \(m_{\text{tangent}} = f'(x_1)\), the derivative evaluated at point \(P(x_1, y_1)\).

Visualizing the Relationship

Curve: \(y = f(x)\)

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Point P(\(x_1, y_1\)) on curve

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Tangent

Gradient = \(f'(x_1)\)

Equation: \(y – y_1 = f'(x_1)(x – x_1)\)

Normal

Gradient = \(-\frac{1}{f'(x_1)}\)

Equation: \(y – y_1 = -\frac{1}{f'(x_1)}(x – x_1)\)

Part 2: Finding Equations of Tangents and Normals

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Step-by-Step Procedure

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General Procedure for Finding Tangents/Normals

Find the derivative: Differentiate \(y = f(x)\) to get \(\frac{dy}{dx} = f'(x)\)

Evaluate at the point: Substitute \(x = x_1\) to find \(m_{\text{tangent}} = f'(x_1)\)

Find the normal gradient: \(m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\) (if \(m_{\text{tangent}} \neq 0\))

Use point-slope form: \(y – y_1 = m(x – x_1)\)
For tangent: \(y – y_1 = f'(x_1)(x – x_1)\)
For normal: \(y – y_1 = -\frac{1}{f'(x_1)}(x – x_1)\)

Simplify: Write in the form \(y = mx + c\) if required

📝 Example 1: Basic Tangent and Normal

Find the equations of the tangent and normal to the curve \(y = x^2\) at the point where \(x = 2\).

Find the point: When \(x = 2\), \(y = 2^2 = 4\), so point is \((2, 4)\)

Find derivative: \(\frac{dy}{dx} = 2x\)

Tangent gradient: At \(x = 2\), \(m_{\text{tangent}} = 2(2) = 4\)

Tangent equation: \(y – 4 = 4(x – 2)\)
Simplify: \(y = 4x – 4\)

Normal gradient: \(m_{\text{normal}} = -\frac{1}{4}\)

Normal equation: \(y – 4 = -\frac{1}{4}(x – 2)\)
Simplify: \(y = -\frac{1}{4}x + \frac{9}{2}\)

📝 Example 2: Polynomial Curve

Find the equation of the tangent to the curve \(y = x^3 – 3x^2 + 2\) at the point where \(x = 1\).

Find the point: When \(x = 1\), \(y = 1^3 – 3(1)^2 + 2 = 1 – 3 + 2 = 0\), so point is \((1, 0)\)

Find derivative: \(\frac{dy}{dx} = 3x^2 – 6x\)

Tangent gradient: At \(x = 1\), \(m_{\text{tangent}} = 3(1)^2 – 6(1) = 3 – 6 = -3\)

Tangent equation: \(y – 0 = -3(x – 1)\)
Simplify: \(y = -3x + 3\)

Special Cases: 1. When \(m_{\text{tangent}} = 0\) (horizontal tangent), the normal is vertical with equation \(x = x_1\)
2. When \(m_{\text{tangent}}\) is undefined (vertical tangent), the normal is horizontal with equation \(y = y_1\)
3. When finding the normal gradient, remember \(m_{\text{tangent}} \neq 0\) (can’t divide by zero)

Part 3: Applications and Problem Types

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CSEC-Style Problems

Type 1: Given Point on Curve

Problem: “Find tangent/normal at point P(\(x_1, y_1\)) on curve \(y = f(x)\)”

Solution: Use standard procedure with given point

Example: Find tangent to \(y = x^2 – 4x\) at \((3, -3)\)

Type 2: Given x-coordinate Only

Problem: “Find tangent/normal where \(x = a\) on curve \(y = f(x)\)”

Solution: First find y-coordinate: \(y_1 = f(a)\), then proceed

Example: Find normal to \(y = \sqrt{x}\) at \(x = 4\)

Type 3: Tangent Parallel/Perpendicular to Given Line

Problem: “Find points where tangent is parallel/perpendicular to line \(y = mx + c\)”

Solution: Set \(f'(x) = m\) (parallel) or \(f'(x) = -\frac{1}{m}\) (perpendicular), solve for x

Example: Find where tangent to \(y = x^3\) is parallel to \(y = 12x\)

Type 4: Finding Points with Given Gradient

Problem: “Find points on curve where gradient equals k”

Solution: Solve \(f'(x) = k\) for x, then find corresponding y

Example: Find points on \(y = 2x^2 – 3x\) where gradient is 5

📝 Example 3: Tangent Parallel to Given Line (Past Paper Style)

Find the equation of the tangent to the curve \(y = x^3 – 3x\) that is parallel to the line \(y = 9x\).

Gradient of given line: \(m = 9\) (from \(y = 9x\))

Derivative of curve: \(\frac{dy}{dx} = 3x^2 – 3\)

Set equal for parallel lines: \(3x^2 – 3 = 9\)

Solve: \(3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = 2\) or \(x = -2\)

Find points:
When \(x = 2\), \(y = 2^3 – 3(2) = 8 – 6 = 2\), point: \((2, 2)\)
When \(x = -2\), \(y = (-2)^3 – 3(-2) = -8 + 6 = -2\), point: \((-2, -2)\)

Tangent equations:
At \((2, 2)\): \(y – 2 = 9(x – 2) \Rightarrow y = 9x – 16\)
At \((-2, -2)\): \(y + 2 = 9(x + 2) \Rightarrow y = 9x + 16\)

📝 Example 4: Finding Point Given Gradient of Normal

The normal to the curve \(y = x^2 + 3x – 4\) at point P has gradient \(-\frac{1}{5}\). Find the coordinates of P.

Relationship: \(m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\)

Find tangent gradient: \(-\frac{1}{m_{\text{tangent}}} = -\frac{1}{5} \Rightarrow m_{\text{tangent}} = 5\)

Derivative of curve: \(\frac{dy}{dx} = 2x + 3\)

Set equal to tangent gradient: \(2x + 3 = 5\)

Solve for x: \(2x = 2 \Rightarrow x = 1\)

Find y-coordinate: \(y = 1^2 + 3(1) – 4 = 1 + 3 – 4 = 0\)

Answer: Point P is \((1, 0)\)

Part 4: Special Cases and Graphical Interpretation

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Visual Understanding

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What Tangents and Normals Look Like

Horizontal Tangent

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• Gradient = 0
• Normal is vertical line

Vertical Tangent

• Gradient undefined
• Normal is horizontal line

Perpendicular Pair

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• Tangent and normal
• Always at 90° to each other

The Differentiation Connection

The derivative \(\frac{dy}{dx}\) gives the gradient function of the curve.

At point \((x_1, y_1)\) on curve \(y = f(x)\):
Gradient of tangent = \(f'(x_1) = \left.\frac{dy}{dx}\right|_{x=x_1}\)

This is why differentiation is essential for finding tangents!

📝 Example 5: Horizontal and Vertical Tangents

For the curve \(y = x^3 – 3x^2\), find:

(a) Points where tangent is horizontal

(b) Equation of normal at point where tangent is horizontal

Find derivative: \(\frac{dy}{dx} = 3x^2 – 6x\)

Part (a) – Horizontal tangent: Gradient = 0, so \(3x^2 – 6x = 0\)

Solve: \(3x(x – 2) = 0 \Rightarrow x = 0\) or \(x = 2\)

Find points:
When \(x = 0\), \(y = 0\), point: \((0, 0)\)
When \(x = 2\), \(y = 2^3 – 3(2^2) = 8 – 12 = -4\), point: \((2, -4)\)

Part (b) – Normal at horizontal tangent:
At \((0, 0)\), tangent gradient = 0 (horizontal), so normal is vertical: \(x = 0\)
At \((2, -4)\), tangent gradient = 0 (horizontal), so normal is vertical: \(x = 2\)

Part 5: Real-World Applications

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Practical Uses of Tangents and Normals

Applications in Various Fields:

Physics: Finding instantaneous velocity (tangent to position-time graph) and acceleration components

Engineering: Determining stresses and strains in materials, optimizing designs

Economics: Marginal cost and revenue (tangent to cost/revenue curves)

Computer Graphics: Rendering curves and surfaces, calculating reflections

Navigation: Calculating optimal paths, especially in aviation and maritime routes

Optics: Law of reflection: angle of incidence = angle of reflection (measured from normal)

📝 Example 6: Application Problem

A ball is thrown and its height (in meters) after t seconds is given by \(h(t) = 20t – 5t^2\).

(a) Find the velocity of the ball after 2 seconds.

(b) Find when the ball reaches its maximum height.

Part (a): Velocity is derivative of height: \(v(t) = h'(t) = 20 – 10t\)

At t = 2: \(v(2) = 20 – 10(2) = 20 – 20 = 0\) m/s

Interpretation: After 2 seconds, velocity is 0 (ball at maximum height)

Part (b): Maximum height when velocity = 0: \(20 – 10t = 0 \Rightarrow t = 2\) seconds

Connection to tangents: Velocity at any time is gradient of tangent to height-time curve at that time

Real-World Insight: In optics, when light reflects off a surface, the angle between the incoming ray and the normal equals the angle between the reflected ray and the normal. This is why normals are so important in physics and computer graphics for calculating reflections.

Comparison Table: Tangent vs Normal

Aspect	Tangent	Normal
Definition	Line touching curve at one point	Line perpendicular to tangent at point of contact
Gradient Relationship	\(m_{\text{tangent}} = f'(x_1)\)	\(m_{\text{normal}} = -\frac{1}{f'(x_1)}\)
Equation Form	\(y – y_1 = f'(x_1)(x – x_1)\)	\(y – y_1 = -\frac{1}{f'(x_1)}(x – x_1)\)
Special Cases	• Horizontal if \(f'(x_1) = 0\) • Vertical if \(f'(x_1)\) undefined	• Vertical if tangent horizontal • Horizontal if tangent vertical
Applications	Instantaneous rate of change, velocity, marginal values	Reflections, perpendicular bisectors, optimization constraints

Common Mistakes to Avoid: 1. Forgetting to find the y-coordinate when only x is given
2. Using wrong sign when finding normal gradient (should be negative reciprocal)
3. Not checking for special cases (horizontal/vertical tangents)
4. Confusing parallel and perpendicular conditions
5. Not simplifying the final equation to required form
6. Forgetting that point must lie on the curve (check by substitution)

Quiz: Test Your Understanding

Tangents and Normals Quiz

Question 1: Find the equation of the tangent to the curve \(y = x^2 – 4x + 3\) at the point where \(x = 3\).

Answer:
1. Point: When \(x = 3\), \(y = 3^2 – 4(3) + 3 = 9 – 12 + 3 = 0\), so \((3, 0)\)
2. Derivative: \(\frac{dy}{dx} = 2x – 4\)
3. Gradient at \(x = 3\): \(2(3) – 4 = 6 – 4 = 2\)
4. Tangent equation: \(y – 0 = 2(x – 3) \Rightarrow y = 2x – 6\)

Question 2: Find the equation of the normal to the curve \(y = 2x^3 – x^2\) at the point \((1, 1)\).

Answer:
1. Derivative: \(\frac{dy}{dx} = 6x^2 – 2x\)
2. Gradient at \(x = 1\): \(6(1)^2 – 2(1) = 6 – 2 = 4\) (tangent gradient)
3. Normal gradient: \(-\frac{1}{4}\)
4. Normal equation: \(y – 1 = -\frac{1}{4}(x – 1)\)
Simplify: \(y = -\frac{1}{4}x + \frac{5}{4}\)

Question 3: Find the points on the curve \(y = x^3 – 6x^2 + 9x\) where the tangent is horizontal.

Answer:
1. Derivative: \(\frac{dy}{dx} = 3x^2 – 12x + 9\)
2. Horizontal tangent when \(\frac{dy}{dx} = 0\): \(3x^2 – 12x + 9 = 0\)
3. Divide by 3: \(x^2 – 4x + 3 = 0\)
4. Factor: \((x – 1)(x – 3) = 0 \Rightarrow x = 1\) or \(x = 3\)
5. Find points:
When \(x = 1\), \(y = 1^3 – 6(1)^2 + 9(1) = 1 – 6 + 9 = 4\), point: \((1, 4)\)
When \(x = 3\), \(y = 27 – 54 + 27 = 0\), point: \((3, 0)\)

Question 4: The tangent to the curve \(y = ax^2 + bx\) at the point \((2, 4)\) has gradient 3. Find the values of a and b.

Answer:
1. Point lies on curve: \(4 = a(2)^2 + b(2) = 4a + 2b\) … Equation (1)
2. Derivative: \(\frac{dy}{dx} = 2ax + b\)
3. At \(x = 2\), gradient = 3: \(2a(2) + b = 3 \Rightarrow 4a + b = 3\) … Equation (2)
4. Subtract Equation (2) from (1): \((4a + 2b) – (4a + b) = 4 – 3 \Rightarrow b = 1\)
5. Substitute \(b = 1\) into Equation (2): \(4a + 1 = 3 \Rightarrow 4a = 2 \Rightarrow a = 0.5\)
6. Answer: \(a = 0.5\), \(b = 1\)

Question 5: Find the equation of the normal to the curve \(y = \frac{1}{x}\) at the point where \(x = 2\).

Answer:
1. Point: When \(x = 2\), \(y = \frac{1}{2}\), so \((2, 0.5)\)
2. Derivative: \(y = x^{-1}\), so \(\frac{dy}{dx} = -x^{-2} = -\frac{1}{x^2}\)
3. Gradient at \(x = 2\): \(-\frac{1}{2^2} = -\frac{1}{4}\) (tangent gradient)
4. Normal gradient: \(-\frac{1}{(-1/4)} = 4\) (negative reciprocal)
5. Normal equation: \(y – 0.5 = 4(x – 2)\)
Simplify: \(y = 4x – 7.5\)

🎯 Key Concepts Summary

Tangent: Line touching curve at one point
- Gradient = \(f'(x_1)\) = derivative at point
- Equation: \(y – y_1 = f'(x_1)(x – x_1)\)
Normal: Line perpendicular to tangent at point
- Gradient = \(-\frac{1}{f'(x_1)}\) (negative reciprocal)
- Equation: \(y – y_1 = -\frac{1}{f'(x_1)}(x – x_1)\)
Special Cases:
- Horizontal tangent (\(m = 0\)) → Vertical normal
- Vertical tangent (\(m\) undefined) → Horizontal normal
Procedure:
- 1. Find point \((x_1, y_1)\) on curve
- 2. Differentiate to get \(f'(x)\)
- 3. Evaluate \(f'(x_1)\) for tangent gradient
- 4. Use negative reciprocal for normal gradient
- 5. Apply point-slope formula
Common CSEC Questions:
- Find equation of tangent/normal at given point
- Find point where tangent has given gradient
- Find where tangent is parallel/perpendicular to given line
- Find horizontal/vertical tangents
- Application problems (velocity, optimization)
Exam Strategy:
- Always find the y-coordinate if only x is given
- Check point lies on curve by substitution
- Show all differentiation steps clearly
- For normals, remember negative reciprocal
- Simplify final equation to required form

CSEC Exam Strategy: When answering tangent/normal questions: (1) Start by finding the point of contact (both coordinates), (2) Differentiate correctly and show your work, (3) Evaluate the derivative at the point for tangent gradient, (4) For normals, use negative reciprocal, (5) Apply point-slope formula, (6) Simplify your answer. Remember: Parallel lines have equal gradients, perpendicular lines have negative reciprocal gradients.